Let $I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$      ...(1)

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x$                       $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$     ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\frac{\pi}{2}}\left\{\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right\} d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x}\right) d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log 1 d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 0 d x$

$\Rightarrow I=0$

Hence, the correct answer is C.